`I = 15 a`
`a = 5.0 cm`
`a = 5 xx 10^(-2) m`
`(mu_0)/(4pi) = 10^(-7) T mA^(-1)`
Magnetic field at the centre of circular loop
`= (mu_0)/(4pi) (2 pi I)/(a)`
Magnetic field at centre of semi-circular loop
`= 1/2 xx (mu_0)/(4pi) (2 pi I)/a`
`= 1/2 xx 10^(-7) xx 2 xx 22/7 xx 15/(5 xx 10^(-2))`
`= 9.42 xx 10^(-5) T`
Magnetic field will be normal to plane of paper but in upward direction.
Magnetic field at the centre of currect carrying arc due to straight portion i.e., due to AB and CD is zero. So at centre , net magnetic field = `9.42 xx 10^(-5) + 0 = 9.42 xx 10^(-5) T`.